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Paper 9709/01
Paper 1
General comments
The paper allowed the vast majority of candidates to show what they had learnt. There were however several
questions which caused most candidates some difficulty (Questions 5, 7 and 9) but allowed the more able
candidates to produce work of a high quality. There were many excellent scripts and not so many scripts at
the lower end. The standard of algebraic manipulation varied considerably from Centre to Centre and the
level of algebra required in Questions 7(i) and 10(ii) proved too much for many candidates. The standard of
presentation was generally pleasing and the majority of candidates followed the rubric by showing their
working in full.
Comments on specific questions
Question 1
4
2
⎛x⎞ ⎛2⎞
C2 ⎜ ⎟ ⎜ ⎟ but this
⎝2⎠ ⎝x⎠
usually followed after writing down the whole expansion. Only a small minority used the general term. It was
pleasing that the binomial coefficient was rarely omitted. A significant number of candidates offered both the
coefficients of x² and x –2 and many of these added these coefficients.
The majority of candidates realised that the coefficient of x² came from the term
Answer: 3
6
3
.
4
Question 2
There was a definite improvement in the way most candidates tackled this question on trigonometric
identities and the majority coped with the algebra required in adding the two fractions and with recognising
the need to use the identity sin 2 x + cos 2 x = 1 . The most common error was to expand (1 + sin x ) 2 as
1 + sin2 x and surprisingly there were a large number of candidates who failed to proceed from
2 + 2 sin x
to the given answer.
cos x(1 + sin x )
Question 3
This proved to be a straightforward question with most candidates obtaining a correct value for the common
difference and proceeding to use the formula for the sum of n terms to obtain a quadratic in n. The most
common errors occurred through incorrect algebra in either forming or solving the quadratic equation.
Answer: 8.
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© UCLES 2008
9709 Mathematics November 2008
Question 4
It was pleasing to see many correct solutions and a very good understanding of the concept of scalar
product. Unfortunately sign errors were very common in part (i), especially the error of expressing PA as
AP . Similarly in part (ii), a large number of candidates used AP . PN instead of PA. PN to calculate angle
APN. A similar number of candidates, having obtained a scalar product of −16, incorrectly assumed that the
modulus of this was needed in order to evaluate the angle.
Answers: (i) −6i − 8j − 6k, 6i + 2j − 6k; (ii) 99.0º.
Question 5
This question was very poorly answered with only a small majority of candidates giving a fully correct
answer. The restriction of ‘where a and b are positive constants’ meant that a + b = 10 and a − b = −2 ,
rather than the more widely found ‘ a + b = −2 ’ and ‘ a − b = 10 ’. Most candidates realised the need in part
(ii) to make cos x the subject, but many of them only gave one answer. The sketch graphs in part (iii) were
mixed with most realising that there was one complete cycle in the range 0° to 360°, but many candidates
produced graphs that were very nearly ‘V’ or inverted ‘V’ shaped shapes.
Answers: (i) 4, 6; (ii) 48.2°, 311.8°.
Question 6
This proved to be a straightforward question that generally produced high marks. Part (i) presented a few
problems with some candidates failing to realise the need to use s = rθ but it was pleasing that most coped
comfortably with using radian measure. Apart from a small minority who failed to halve the angle found in
part (i), parts (ii) and (iii) generally produced correct answers.
Answers: (i) 1.8 radians; (ii) 6.30 cm; (iii) 9.00 cm².
Question 7
Part (i) caused lot of problems with at least a third of all attempts failing to recognise the need to obtain two
simultaneous equations in x and r. Of the others, most attempted to make r the subject, but the algebra
⎛ 40 − 2 x ⎞
required in squaring ⎜
⎟ defeated at least a half of them. Part (ii) was well answered and it was
⎝ π
⎠
pleasing to see that most of the candidates who were unable to answer part (i) realised that part (ii) was still
accessible. Nearly all candidates realised the need to differentiate and to set the differential to 0.
Answer: (ii) 11.2.
Question 8
This proved to be a source of high marks and there were a large number of perfectly correct solutions. In
part (i) most candidates differentiated correctly and used the formula m1m2 = −1 to find the gradient of the
normal, though many weaker candidates failed to realise the need to express this as a number prior to
finding the equation of a straight line. Parts (ii) and (iii) were usually correctly answered.
Answers: (ii) (−8, 6); (iii) 11.2.
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© UCLES 2008
9709 Mathematics November 2008
Question 9
In part (i) a minority of solutions found the answer directly from making x the subject and finding the area by
using
∫ x dy .
Of those using
∫ y dx , failure to obtain the correct multiplying constant in the integration of
1
(3 x + 1) 2 or to subtract the resulting area from the area of a rectangle (2) meant that there was only a small
proportion of these candidates obtaining a correct answer. Similarly in part (ii), many candidates either
incorrectly attempted to find the volume about the y-axis, or failed to subtract the volume about the x-axis
from the volume of a cylinder. There were very few correct answers for part (iii) with most of them coming
from use of the formula for tan(A − B) (a P3 topic). Only a small minority recognised that the gradient of the
tangent can be equated to tan θ, where θ is the angle made with the x-axis.
Answers: (i)
4
; (ii) 4.71; (iii) 19.4°.
9
Question 10
The general response to this question was mixed. Parts (iii) and (iv) were very well answered; parts (i) and
(ii) much less so. In part (i), most candidates managed to sketch the graph of y = f(x) but many failed to
realise that the graph of y = f−1(x) was the reflection of the graph of y = f(x) in the line y = x. In part (ii), nearly
all candidates obtained a correct expression for gf(x) but failed to cope with the algebra of simplifying
6(3 x − 2) − (3 x − 2) 2 to obtain a correct quadratic. Those proceeding to differentiate to obtain a correct value
of x at the stationary value and hence to prove the required result fared much better than those attempting to
complete the square. A very large number of these candidates changed the sign of the quadratic but failed
either to complete the square correctly or to reverse the sign at a later time. Part (iii) presented very few
problems and it was pleasing that most candidates realised the need to use the answer to part (iii) in order to
find the inverse of h. Several candidates however left the answer in terms of y instead of x.
Answers: (ii) −9x2 + 30x − 16; (iii) 9 − (x − 3 ) ; (iv) 3 + 9 − x .
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© UCLES 2008
9709 Mathematics November 2008
MATHEMATICS
Paper 9709/02
Paper 2
General comments:
Candidates generally showed poor understanding of the basic rules and results of the calculus in Questions
5, 6, and 8(ii), (iii).
Many marks were lost by candidates using the correct methods but spoiling promising solutions by careless
mistakes; often seen, for example, was the error ‘2a + 26 = 0, so a = +13’ in Question 2(i). Such errors
jeopardise later parts of the working.
Where exact answers were requested, marks were often lost by use of approximate values, e.g. in Question
4(i) and the later stages of Question 5. Candidates are advised to read the questions more carefully.
Standards of neatness and presentation were very high. There was no evidence of candidates having too
little time to complete the paper.
Candidates are strongly advised to prepare themselves by working through, and studying carefully, previous
papers.
Comments on individual questions
Question 1
Almost everyone squared each side to obtain a quadratic equation (or inequality) in x. Others failed to
proceed beyond x = −3 or x = +1. Although the question was generally successfully attempted, the final
inequality was often incorrectly formed. Candidates are advised to check if the value x = 0 satisfies the
original inequality; if it does, then it must belong to the final solution set.
Answer: −3 < x < 1.
Question 2
(i)
(ii)
Candidates invariably noted that p(–2) = 0, but solutions were often marked by a variety of sign
errors.
A correct value for a was usually followed by a correct solution here, but many candidates
unnecessarily solved the equation p(x) = 0.
Answers: (i) −13; (ii) (x + 2)(2x + 1)(x – 3).
Question 3
Many candidates failed to attempt this question. Here ln y (not y itself) is linear in x, thus ln y = ln A – x ln b,
1
with ln A being the intecept on the vertical axis and – ln b being the gradient (= − ) of the line.
4
Answers: 3.67, 1.28.
Question 4
(i)
The equation was often given as sin x cos 60° + cosx sin 60° = 2 (cos x cos 60° + sin x sin 60°) or
the right-hand side of the equation as (2 cos x × 2 cos 60° – 2 sin x × 2 sin 60°). Many good
solutions stopped short of a final answer.
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© UCLES 2008
9709 Mathematics November 2008
(ii)
Around half of all candidates derived an equation tan x = 3 3 or tan x = 0, instead of the correct
1
. The second correct solution, in the negative second quadrant, was only rarely
form tan x =
3 3
found.
Answer: (ii) 10.9°, −169.1°.
Question 5
1
, but few could obtain a multiple of ln(2x + 1) on integrating the
x
ln a
, or
second term. Often a promising solution was ended by use of a false rule of the form (ln a – ln b) =
ln b
by using approximate values for ln 2, ln 3 and ln 5.
Almost everyone correctly integrated
Question 6
Alarmingly, most candidates did not realize that the derivative of a product, y = f(x).g(x) takes the form
xe − x
. Those obtaining a correct first derivative
(fg′ + gf′ ). Thus y′ was given as a single term, usually as −
2
often struggled to obtain a correct form for y″. A common error was to solve the equation y′ = 0. The correct
derivatives were e
1
− x
2
1
(1 −
− x⎛ x
1
⎞
x ) and e 2 ⎜ − 1⎟ .
2
⎝4 ⎠
Answer: (4, 4e –2).
Question 7
(i)
Graphs were generally poor. Many candidates drew that of y = (2 – 2x) as a curve, rather than a
line, and few had the intersection as lying between x = 0.5 and x = 1.
(ii)
Candidates needed to form a function f(x) = ± (2 − 2x − cos x), to evaluate f(0.5) and f(1.0) and to
note that differ in sign.
(iii)
Many candidates used cos x = 2 – 2x in an attempt to derive the iterative formula instead of letting
1
n tend to infinity in the iterative formula and obtaining x = 1 – cos x.
2
(iv)
Iteration was usually successful, though many solutions were not rounded to 2 decimal places.
Those who erroneously took x to be in degrees (rather than radians) obtained iteratives all equal to
0.5000.
Answer: (iv) 0.58.
Question 8
(i)
A roundabout argument leading nowhere was common in part (a). In part (b), a common serious
1 + sin x
.
error was to cancel the term ‘sin’ in the expression
1 − sin2 x
(ii)
Differentiation was often poor and not based on use of D{(f(x))n} = n f(x)n – 1. f′ (x).
(iii)
Few correct integrations were performed. The results of parts (i)(b) and (ii) were required. Many
candidates wrongly substituted the limits into the integrand.
Answer: (iii) √2.
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© UCLES 2008
9709 Mathematics November 2008
MATHEMATICS
Paper 9709/03
Paper 3
General comments
The standard of work on this paper varied considerably and resulted in a wide spread of marks. Candidates
found this to be one of the more challenging papers set on this syllabus. Nevertheless candidates appeared
to have sufficient time to attempt all questions and no question seemed unduly difficult. The questions or
parts of questions that were done well were Question 1 (logarithms), Question 2 (binomial expansion),
Question 4 (parametric differentiation) and Question 5(i) (algebra). Those that were done least well were
Question 6(ii) (trigonometric equation), Question 8 (differential equation), and Question 10 (complex
numbers).
In general the presentation of work was good but there were still candidates who presented their work in a
double column format. This makes marking difficult for Examiners and it would be helpful if Centres could
continue to discourage the practice.
Where numerical and other answers are given after the comments on individual questions, it should be
understood that alternative forms are often acceptable and that the form given is not necessarily the sole
‘correct answer’.
Comments on individual questions
Question 1
This was generally well answered. Some candidates incorrectly assumed ln(x + 2) = ln x + ln 2, or
ln(x + 2) = ln x × ln 2, but most removed logarithms correctly, reaching the equation x + 2 = e 2 x , or
equivalent. A significant number failed to solve this correctly for x.
Answer: 0.313 .
Question 2
Most candidates wrote down the correct relevant unsimplified terms of the expansion of
(1 − 2 x ) , though
1
1
or −2 instead of
. Whatever the expansion, most made sure that the
2
2
product with (1 + x) contained all the necessary terms. Some confined themselves to finding the coefficient
of the term in x 2 instead of the complete expansion up to this term.
some took the index to be −
Answer: 1 −
3 2
x .
2
Question 3
The initial differentiation was usually done well using the quotient or product rule. In some cases the quotient
rule was misapplied, e.g. by taking u = cos x instead of u = e x , The presence of e x in the equation
e x cos x + e x sin x = 0 or e x sec x + e x sec x tan x = 0 unsettled some candidates but the main source of
1 1
error was the inability to find the root of tan x = −1 in the interval ( − π , π ) .
2 2
1
Answer: − π or –0.785 radians.
4
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© UCLES 2008
9709 Mathematics November 2008
Question 4
This was generally very well answered. Candidates who used the double angle formulae after differentiating
tended to be more successful than those who substituted at the beginning.
Question 5
There were many correct answers to part (i). In part (ii) the correct final answer appeared regularly, but was
only very rarely justified properly. Some candidates showed and stated that 2 x 2 − 3 x + 3 had no real zeroes
but almost always they omitted to either show that it was positive for all x or alternatively examine the sign of
1
p(x) when x took a value other than the sole critical value − .
2
1
Answers: (i) 3; (ii) x < − .
2
Question 6
Part (i) was usually done well. In part (ii) candidates were expected to convert the given equation to the
form 13 sin( 2θ + α ) = 11 . However common errors were to treat the argument as 2(θ + α), θ + α, or θ + 2α.
Some worked with x + α as argument but did not complete the solution by using x = 2θ. Those with a sound
approach usually found a root in the interval 0° < θ < 90° but less frequently found the second one in the
interval 90° < θ < 180°. Moreover, spurious roots quite commonly appeared in the solution.
Answers: (i) 13 sin(x + 67.38°); (ii) 27.4°, 175.2°.
Question 7
(i)
This was generally done well. The normal vector 2i − j − 3k was sometimes miscopied as
2i − j + 3k but the correct processes seemed to be well known. Errors in nominating the final acute
angle were quite common, with 122.3° or even 57.7° being followed by 32.3° as final answer.
(ii)
A minority clearly did not know how to tackle this type of problem. However the majority attacked it
with a variety of appropriate methods. Careful checking might have saved the loss of marks caused
by algebraic and numerical slips.
Answers: (i) 57.7° or 1.01 radians; (ii) r = 2i – k + λ(4i –7j +5k).
Question 8
This question was poorly answered. In part (i) many candidates, having obtained
dV
= 4h 2 , failed to
dh
dV
dh
= 20 − kh 2 . Often
was taken to represent the rate of flow of
dt
dt
liquid. Candidates who merely verified that the boundary conditions satisfied the given equation scored zero.
interpret the question as implying that
The work in part (ii) was also poor. Many candidates seemed to think that verifying the identity for one value
of h was enough. There were many incorrect attempts at separating variables in part (iii). The minority who
20h 2
separated correctly often failed to use the identity given in part (ii) and made errors in converting
100 − h 2
to an integrable form.
⎛ 10 + h ⎞
Answer: (iii) t = 100 ln⎜
⎟ − 20h .
⎝ 10 − h ⎠
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© UCLES 2008
9709 Mathematics November 2008
Question 9
1
(i)
x
Most tried to integrate by parts but often made errors in the integration of e 2 . Those who worked
1
a
correctly seemed to find the final steps from a 2 (2a − 4) = 2 to the given answer difficult or in
some cases impossible.
(ii)
Almost all candidates attempted to make sketches of the graph of an exponential function and the
appropriate corresponding straight line. Some of the former had the wrong curvature, and some
only showed the part of the curve for x > 0. Even when both graphs were correctly sketched and
provided sufficient evidence for a conclusion to be drawn, many failed to state or indicate that the
existence of just one point of intersection implied that the equation had only one root.
(iii)
This was satisfactorily answered. There were some candidates who seemed to believe that a
statement involving ‘positive’ and ‘negative’ was sufficient. However the majority made clear the
function they were considering and calculated values as requested, before stating what the change
in sign meant.
(iv)
Since no iterative formula was given, this part began by testing whether candidates could produce
−1x
one. Those who stated, or were clearly working with, the formula x n +1 = 2 + e 2 n answered the
question well. However some of the candidates who did not state a formula produced sequences
that seemed to Examiners to be unrelated to the equation in part (i).
Answer: (iv) 2.31 .
Question 10
Parts (i) and (ii) were generally well answered. Part (iii) was hardly ever correctly answered. There seemed
a widespread belief that the moduli of the three complex numbers were the lengths of the sides of the
triangle whose vertices represented them. Candidates rarely took the results of part (ii) to mean that the
2
three vertices were equidistant from the origin and that two pairs subtended π at the origin. In part (iv)
3
4 + 2i
were the required numbers. Irrelevant work finding the square roots
hardly any saw that (4 + 2i)w and
w
of 4 + 2i was often seen instead.
Answers: (i) 1,
2
2
2
π (or 2.09 radians); (ii) R, θ + π ; R, θ − π ;
3
3
3
(iv) − (2 + 3 ) + (2 3 − 1)i ,
− (2 − 3 ) − (2 3 + 1)i .
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© UCLES 2008
9709 Mathematics November 2008
MATHEMATICS
Paper 9709/04
Paper 4
General comments
The work of candidates was generally good and well presented.
candidates in Questions 1, 2, 3, 5 and 6.
High marks were scored by many
Notwithstanding the good work undertaken by candidates, it was disappointing to see that some elementary
errors were made with considerable frequency. These included: giving answers for the components of one
of the forces instead of those for the resultant of the two forces in Question 1(ii); using the answer in
1
Question 3(i) in part (ii), where it has no relevance; calculating the distance OA from 0.5 × 7 or from g 7 2
2
in Question 4; assuming the time for stage 1 = the time for stage 3 = 200 s in Question 6(ii); obtaining
0.3t 2
0.3t 3
from
dt in Question 7(ii).
2
3
∫
Comments on specific questions
Question 1
(i)
This was poorly attempted with many candidates misunderstanding what was required. The most
common wrong answer in part (a) was –8cos θ and this was usually accompanied by the correct
answer 8sin θ in part (b). These are answers for the components of the force of magnitude 8, and
the correct answer in part (b) is almost certainly fortuitous in most such cases.
(ii)
This part of the question was well attempted. Most candidates used R2 = X2 + Y2 and solved the
resulting equation for cos θ . Almost all such candidates using this approach scored all three
marks, including many who failed to score both marks in part (i). A few candidates solved the
simultaneous equations X = 8cos φ , Y = 8sin φ for cos θ , where φ is the angle the resultant of the
two given forces makes with the ‘x-axis’. However a much greater number of candidates assumed
implicitly from the outset that this angle is θ , not distinguishing it from the θ of the question. Such
X
Y
candidates used 8cos θ =
or tan θ =
, both of which lead fortuitously to the correct answer,
R
X
without justifying the use of θ for φ . Another very common approach was to use a triangle of
forces method. This included either the use of the cosine rule or, less commonly, recognising the
1 10
triangle as being isosceles and hence cos θ =
×
. Finally a formula method involving
2 8
c2 = a2 + b2 + 2abcos C was frequently used, where c is the magnitude of the resultant of two
forces of magnitudes a and b, and the angle between their directions is C.
Answers: (i)(a) 10 − 8 cos θ , (b) 8 sin θ .
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© UCLES 2008
9709 Mathematics November 2008
Question 2
This question was well attempted with many candidates scoring high marks. Errors made by candidates
included taking the normal reaction force as 20g, making sign errors, and omitting the weight component in
part (i). Although it was expected that ‘a force acts on the block parallel to a line of greatest slope of the
plane’ would be well understood, a significant number of candidates reacted incorrectly to the words ‘up’ and
‘down’. Such candidates included the normal reaction force on resolving forces in part (i), and the weight
instead of its component in part (ii).
Answers: (i) 97.8 N; (ii) 28.3 N.
Question 3
This was the best attempted of the early questions and a large proportion of candidates scored all four marks
in part (i). The given answer in part (ii) was achieved by almost all of the candidates, but the level of
18000
ranged from non-existent to excellent.
understanding of why the given lower bound arises from
900
Answer: (i) 0.15 ms−2.
Question 4
This question was poorly attempted. Very many candidates failed to find the correct value for the distance
OA, often inappropriately applying the principle of conservation of energy to obtain 0.0125 metres. Many
candidates used ‘work done = force × distance’, despite the instruction to use energy. Those who did use
energy included many who incorrectly used work done = PE, or work done = KE, or work done = PE – KE.
Answer: 2820 J.
Question 5
This was the best attempted question in the paper. Many candidates started by writing down equations
obtained by applying Newton’s second law to A and to B. This leads to a pair of simultaneous equations in
three unknowns, a, T and m. Most such candidates were able to find the route to a which is independent of
T and m and, after finding the value of a, were able to find T and m.
Answers: (i)(a) 2.5 ms−2, (b) 3.75 N; (ii) 0.3.
Question 6
Almost all candidates scored both of the available marks in part (i), but part (ii) was poorly attempted.
20000
), 30 ms–1 (from 0 + 0.15 × 200),
Frequently occurring wrong answers included 20 ms–1 (from
1000
20000
0+v
=
), and 25 ms–1 using the assumption that the times for the first and third stages
40 ms–1 (from
1000
2
are equal.
In part (iii) many candidates used correct methods but some assumed that the time for stage 1 = the time for
stage 3 = 200 s.
Answers: (ii) 25; (iii) 2920 m.
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© UCLES 2008
9709 Mathematics November 2008
Question 7
Many candidates scored full marks in part (i). Most candidates realised the need to integrate in part (ii) and
many candidates did so twice to obtain an expression for the distance x(t). Integration was usually executed
accurately. However some candidates integrated only once to obtain what they declared to be an expression
for x(t).
Few candidates dealt with the constants of integration or with the evaluation at the final stage correctly. The
first constant of integration was more frequently found to be 0.0375 (from 5 – 10 × 0.5 + 0.15 × 0.52), or zero,
than the correct value of 5. Similarly the second was frequently found to be a value based on x(0.5) = 0
instead of the correct value of 0 or 1.25 (depending on whether the x(t) represents the displacement from A
or from O).
At the final stage more candidates substituted t = 3 than substituted the appropriate value of t = 2.5. Almost
all candidates who evaluated a definite integral at the second stage found
∫
3
0 .5
v (t ) dt instead of
2 .5
∫ v (t ) dt .
0
Answers: (i) 5 ms−1, 0.5 s; (ii) 44.2 m.
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© UCLES 2008
9709 Mathematics November 2008
MATHEMATICS
Paper 9709/05
Paper 5
General comments
The more able candidates were able to score well, but unfortunately some candidates scored low marks and
were clearly not ready for the examination at this level.
The work from some candidates was poorly presented and often difficult to read. However more candidates
are drawing clear diagrams to help them with their solutions.
Only a few candidates used premature approximation and rounded to less than 3 significant figures. The
question paper clearly states that g = 10 should be used and candidates rarely used g = 9.8 or 9.81.
The more able candidates scored well on Questions 1 to 4. Questions 5, 6 and 7 proved to be more of a
challenge with Question 5 found to be the most difficult question on the paper, requiring candidates to use
trigonometry to work out various distances.
Some candidates had a limited understanding of how to take moments.
Comments on specific questions
Question 1
Some candidates tried to use an energy equation which was not productive. At times x = 6 was used for the
x
. Sometimes T = 8a only was seen instead of T – 8g = 8a.
extension in T = λ
l
Answer: 4 ms–2.
Question 2
(i)
Unfortunately the centre of gravity of the cone from the base was sometimes taken as
(ii)
Most candidates obtained α = 39.8, but quite a number found the complement, 50.2.
h
3h
or
.
3
4
Answers: (i) 48; (ii) 39.8.
Question 3
(i)
Most of the candidates gained full credit.
(ii)
Often the integration was correctly done. On occasions either v or v2 was seen instead of
Some candidates did not know how to integrate 16e–x.
v2
.
2
Answer: (ii) 5.33 ms–1.
12
© UCLES 2008
9709 Mathematics November 2008
Question 4
ω
first and then using v = r ω .
(i)
Some candidates made part (b) rather long winded by finding
(ii)
Again some candidates made this part long winded by first finding v and then using
ω
=
v
.
r
Sometimes candidates used 2.5 – 0.671 = 0.15 × 0.2 ω 2 or 2.5 = 0.25 × 0.2 ω 2 .
Answers: (i)(a) 1.5 N, (b) 0.671 N; (ii) 9.13 rad s–1.
Question 5
This was found to be the most difficult question on the paper.
(i)
This part needed a really good clear diagram to help the candidate find the distances and the
forces needed to solve the problem. Most candidates attempted to take moments about D. Often
0.8T = 350 × 0.6cos 20° was seen instead of 0.8T = 360 × 0.6cos 20° − 350 × 0.4sin20°. Some
candidates attempted this part by simply trying to resolve in numerous directions. This approach
was not productive.
(ii)
Candidates made a better attempt at this part of the question. Candidates tried to find F and R and
then used F = μ R. Unfortunately F and R were often incorrectly calculated.
Answer: (ii) 0.424.
Question 6
(i)
Most candidates attempted to use an energy equation.
Many candidates often found
v2 = 40 + 10x – 5x2 but then simply stated that v2 = 45 – 5(x − 1)2. This was too big a step to take
since the answer is given on the question paper.
(ii)
Too many candidates found x = 4 but then omitted to add 4 to give the distance at the lowest point.
(iii)
This was generally well done. Some candidates substituted x = 0 instead of x = 1 into
v2 = 45 – 5(x − 1)2.
Answers: (ii) 8 m;
(iii) 6.71 ms–1.
Question 7
(i)
Quite a number of candidates did not express x and y in terms of V and t. These candidates simply
quoted the equation of the trajectory and substituted θ = 60° into it.
(ii)
This part of the question was often done correctly. Some candidates had difficulty in manipulating
the expression to make V the subject.
(iii)
Only a handful of candidates differentiated the answer in part (i) to find
dy
40 x
= 3 − 2 and then
dx
V
went on to find the required direction of motion of P at the instant it passes through A.
Answers: (i)
x = Vt cos 60° , y = 3 x −
20 x 2
; (ii) 29.7; (iii) 55.3° downward from the horizontal.
V2
13
© UCLES 2008
9709 Mathematics November 2008
MATHEMATICS
Paper 9709/06
Paper 6
General comments
The paper proved accessible to most candidates. There was no problem with shortage of time, and almost
everybody attempted questions in numerical order. It was pleasing to see that the majority of candidates
worked with an appropriate number of significant figures and therefore did not lose any accuracy marks.
However, those who worked with 3 significant figures, i.e. with premature rounding, invariably lost the final
accuracy mark because their answer was not accurate to 3 significant figures.
Comments on specific questions
Question 1
( Σ)
x 2 . This question was
This was usually answered well. Most errors arose from confusion over Σx 2 and
one example where premature rounding, which gave an answer of 4.56 instead of 4.57, was evident.
Answers: 38.4 mm, 4.57 mm.
Question 2
There were many good solutions obtaining full marks. Some candidates tried to use the binomial distribution
which gained no marks as it is not an approximation, which the question asked for. There were the usual
number of errors concerning standard deviation and variance, and continuity corrections. It is pleasing to
see candidates are using diagrams more often to determine the tail.
Answer: 0.652.
Question 3
Generally, candidates who managed Question 2 successfully also coped well with this normal distribution
question. Candidates who used a continuity correction here did not gain the method mark. There was also
some confusion over 1 − Φ( z ) and Φ( 1 − z ) . Candidates who used 1 – 1.22 did not gain credit for method.
Some candidates found the negative numbers difficult to manage, and had –15.1 to the right of 0 on the
normal curve in part (i), whilst in part (ii) 0° C was to the right of the mean. Incorrect answers of 0.0277,
0.0278, etc. were seen due to premature rounding and again these answers lost the final mark as they are
not accurate to 3 significant figures.
Answers: (i) 0.0276; (ii) 7.72.
Question 4
This question was poorly answered. Most candidates gave an answer of 12! to the first part, failing to
appreciate that some of the houses were the same. For part (ii) many candidates could not determine the
number in each group, and if they did, they invariably added instead of multiplied for the final probability. In
part (iii) many errors arose from candidates using 10C2 instead of 7C2.
Answers: (i) 831 600; (ii) 900; (iii) 126.
14
© UCLES 2008
9709 Mathematics November 2008
Question 5
This question was generally well answered. There were some instances of wrong stems of 1, 2, 3, etc. or
100, 110, 120, etc. and some leaves were not placed in columns but were spaced out. In part (ii) a few
candidates did not add 1 to the number of items (15) and thus found the quartiles wrongly. These answers
could have been read off and written down in 3 lines but many candidates took over one side of working to
arrive at the answers. Almost everybody knew what a box-and-whisker plot was, but nearly everybody lost
credit because ‘pulse rate’ or ‘beats per minute’ was not seen on the graph. It is essential to put the units in
when using statistical illustrations. A few drew whisker lines right through the box. Boxes should be empty, in
the order of 1 cm high (not 5 or 6 as was often seen) and the ends of whiskers should be shown by a mark of
some sort. A ruler should be used.
Answers: (ii) 125, 115, 145.
Question 6
Many candidates scored no marks for this question. A three-stage tree diagram proved problematic for
many. Some solutions had an incorrect part (i) but then recovered in parts (ii) and (iii), whilst others had a
correct part (i) but only had two probabilities in later parts. Overall there was very little correlation in this
question between which of parts (i), (ii) or (iii) candidates got correct. Part (iv) however was universally
poorly done with most candidates only finding one option for the numerator instead of two, and those who
did find two options and summed them correctly then failed to divide by their answer to part (iii). Able
candidates scored full marks and this question served as a good discriminator.
Answers: (ii) 0.224; (iii) 0.392; (iv) 0.633.
Question 7
Many good candidates answered this question well. The weaker candidates failed either to find the
probability of throwing an odd number or to recognise a binomial situation, or both. Having negotiated both
those hurdles, errors then occurred in finding the probability of at least 7 odd numbers. Candidates found
P(7), P(8), 1 – any combination of the two, and so on. Parts (ii) and (iii) were straightforward and almost
everybody picked up some marks here. Part (iv) needed a little more understanding which was lacking in
many of the weaker candidates.
Answers: (i) 0.195; (ii) 2,
(iii)
1
2
5
4
4
4
4
8
4
; 4,
; 6,
; 7,
; 8,
; 9,
; 10,
; 11,
, 12,
;
36
36
36
36
36
36
36
36
36
26
5
; (iv)
.
3
9
15
© UCLES 2008
9709 Mathematics November 2008
MATHEMATICS
Paper 9709/07
Paper 7
General comments
This paper proved to be reasonably straightforward, with candidates able to demonstrate and apply their
knowledge in the situations presented. There were many good scripts, with few candidates appearing totally
unprepared for the paper. In general, candidates scored well on Questions 7 and 4(i), (ii), whilst Questions
1, 2 and 4(iii) proved more demanding. It was pleasing to note that Question 5, which required knowledge
of Type I and Type II errors, was reasonably well attempted by some candidates. In previous papers, this
has not always been the case.
Accuracy, as always, caused loss of marks for some candidates; there were a few cases of candidates not
adhering to the accuracy required, either by rounding too early in the question or by giving a final answer to
only 2, or even 1, significant figures. This was particularly noticeable in Question 5. Lack of time did not
seem to be a problem for candidates.
Detailed comments on individual questions follow. Whilst the comments indicate particular errors and
misconceptions, it should be noted that there were also many very good and complete answers.
Comments on specific questions
Question 1
This question was only well attempted by a minority of candidates. Some candidates gave wholly incorrect
and irrelevant explanations, such as considering the musical talent of the children. In part (i) some
candidates realised that the method was unsatisfactory, but did not give a full enough answer to obtain all
available marks. A statement such as ‘not all children have the same chance of being chosen’, whilst a true
statement, does not explain why they do not have equal chance and is not related to the question in any
way. Similarly in part (ii) many candidates did not give a detailed enough answer to explain a full and valid
method.
Question 2
Weaker candidates did not score well on this question. Common errors included using a one-tail test, using
15.7
when finding the test statistic. Some candidates failed to find the average time of the
15.7 rather than
3
given sample of size 3 and attempted to find 3 separate test statistics, showing a lack of understanding of
what was required. Other errors included incorrect comparisons (e.g. incorrect z critical, or even a
comparison of an area with a z-value) and in some cases no comparison was shown to justify a final
conclusion. Any final statement in a hypothesis test such as this must be justified by a relevant comparison.
Many candidates wrote their null and alternative hypotheses as H0 = 42 and H1 ≠ 42, omitting μ. It was
disappointing to find that candidates, who sometimes did the rest of the question successfully, lost marks in
this way.
Answer: Teacher’s estimate can be accepted.
Question 3
Many candidates were able to offer a good solution to this question, and a good attempt was often made
even by weaker candidates. The most common error, as is always the case on this type of question, was an
2
2
2
2
2
2
incorrect variance obtained from 2 x 1.6 + 10 x 0.4 rather than the correct variance of 2 × 1.6 + 10 × 0.4
Answer: 0.350.
16
© UCLES 2008
9709 Mathematics November 2008
Question 4
As mentioned above, candidates found parts (i) and (ii) of this question straightforward. There were few
consistent errors other than the usual confusion between biased and unbiased estimates, and the two
different formulas for the unbiased estimate of the variance. The most successful candidates used the
formula given in the formula list. Many candidates found the correct confidence interval; the most common
errors noted were incorrect z-values or an incorrect formula.
Part (iii), however, was poorly attempted and omitted completely by many candidates. It was surprising to
note, that whilst candidates could find a confidence interval, they were unable to interpret what it meant.
Answers: (i) 4.27, 0.00793; (ii) (4.25, 4.29); (iii) 9.
Question 5
It was pleasing to note that some candidates correctly identified the probability of a Type I and a Type II error
in the given situation. There were still cases where candidates quoted what was meant by these errors, and
were unable to go any further and relate this knowledge to the question, but only weaker candidates did not
actually make any attempt at finding probabilities. In parts (i) and (ii) many candidates reversed the
probabilities finding P(0) instead of P(1 or more) in part (i) and P(1 or more) instead of P(0) in part (ii). Many
candidates correctly used a Poisson approximation in part (iii), though the correct value of λ (0.288) was not
always found.
Answers: (i) 0.0202; (ii) 0.972; (iii) 0.0311.
Question 6
Most candidates correctly found the answer to part (i), though some incorrectly felt the answer required
rounding to the nearest whole number. Parts (ii) and (iii) were not always well answered. Some candidates
did not realise the need to multiply two Poisson probabilities in part (ii). In part (iii), whilst many candidates
calculated 1 – P(0), some did not give this as a final answer, and some did not use the correct value of λ.
Similarly in part (iv) candidates did not always calculate a new value of λ and merely used the value of 1.8.
Answers: (i) 1.15; (ii) 0.216; (iii) 0.784; (iv) 0.776.
Question 7
This was a well attempted question, even by weaker candidates. Most candidates correctly showed that k
3
, and there were few cases of candidates showing insufficient working. On the whole, attempts at
was
14
integration were good, though some candidates used unnecessarily long methods (integration by parts).
There was an occasional confusion between part (ii) requiring the mean and part (iii) requiring the median.
On the whole errors were mainly calculation errors, though an error of limits in part (iii) was quite common
with candidates integrating from ‘0 to m’ rather that from ‘1 to m’. Many candidates used a correct method
for part (iv); once again errors were mainly from incorrect calculations rather than from errors in method.
Answers: (ii) 2.66 hours; (iii) 2.73 hours; (iv) 0.0243.
17
© UCLES 2008